EXERCISE 4.1Solve The Following Quadratic Equations:1. $(x-1)(x-2)=0$2. $(x-4)(x+5)=0$3. $(3-x)(2-x)=0$4. $(11+m)(3-m)=0$

Introduction

Quadratic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. In this article, we will focus on solving quadratic equations of the form $(x-a)(x-b)=0$, where $a$ and $b$ are constants. We will use four examples to illustrate the steps involved in solving these types of equations.

What are Quadratic Equations?

A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable (in this case, $x$) is two. The general form of a quadratic equation is:

$$ax^2 + bx + c = 0$$

where $a$, $b$, and $c$ are constants, and $a$ cannot be zero.

Solving Quadratic Equations: A Step-by-Step Guide

To solve a quadratic equation of the form $(x-a)(x-b)=0$, we need to follow these steps:

1. Factor the equation: Factor the quadratic expression into two binomial factors.
2. Set each factor equal to zero: Set each binomial factor equal to zero and solve for $x$.
3. Solve for $x$: Solve the resulting linear equations to find the values of $x$.

Example 1: $(x-1)(x-2)=0$

Let's use the first example to illustrate the steps involved in solving a quadratic equation.

Step 1: Factor the equation

The given equation is already factored into two binomial factors: $(x-1)$ and $(x-2)$.

Step 2: Set each factor equal to zero

Set each binomial factor equal to zero:

$$x-1=0 \quad \text{and} \quad x-2=0$$

Step 3: Solve for $x$

Solve the resulting linear equations to find the values of $x$:

$$x=1 \quad \text{and} \quad x=2$$

Therefore, the solutions to the equation $(x-1)(x-2)=0$ are $x=1$ and $x=2$.

Example 2: $(x-4)(x+5)=0$

Let's use the second example to illustrate the steps involved in solving a quadratic equation.

Step 1: Factor the equation

The given equation is already factored into two binomial factors: $(x-4)$ and $(x+5)$.

Step 2: Set each factor equal to zero

Set each binomial factor equal to zero:

$$x-4=0 \quad \text{and} \quad x+5=0$$

Step 3: Solve for $x$

Solve the resulting linear equations to find the values of $x$:

$$x=4 \quad \text{and} \quad x=-5$$

Therefore, the solutions to the equation $(x-4)(x+5)=0$ are $x=4$ and $x=-5$.

Example 3: $(3-x)(2-x)=0$

Let's use the third example to illustrate the steps involved in solving a quadratic equation.

Step 1: Factor the equation

The given equation is already factored into two binomial factors: $(3-x)$ and $(2-x)$.

Step 2: Set each factor equal to zero

Set each binomial factor equal to zero:

$$3-x=0 \quad \text{and} \quad 2-x=0$$

Step 3: Solve for $x$

Solve the resulting linear equations to find the values of $x$:

$$x=3 \quad \text{and} \quad x=2$$

Therefore, the solutions to the equation $(3-x)(2-x)=0$ are $x=3$ and $x=2$.

Example 4: $(11+m)(3-m)=0$

Let's use the fourth example to illustrate the steps involved in solving a quadratic equation.

Step 1: Factor the equation

The given equation is already factored into two binomial factors: $(11+m)$ and $(3-m)$.

Step 2: Set each factor equal to zero

Set each binomial factor equal to zero:

$$11+m=0 \quad \text{and} \quad 3-m=0$$

Step 3: Solve for $x$

Solve the resulting linear equations to find the values of $x$:

$$m=-11 \quad \text{and} \quad m=3$$

Therefore, the solutions to the equation $(11+m)(3-m)=0$ are $m=-11$ and $m=3$.

Conclusion

In this article, we have discussed how to solve quadratic equations of the form $(x-a)(x-b)=0$. We have used four examples to illustrate the steps involved in solving these types of equations. By following the steps outlined in this article, you should be able to solve quadratic equations with ease.

Tips and Tricks

• Make sure to factor the equation correctly before setting each factor equal to zero.
• Check your solutions by plugging them back into the original equation.
• If you are having trouble solving a quadratic equation, try using the quadratic formula.

Practice Problems

Try solving the following quadratic equations:

1. $(x+2)(x-3)=0$
2. $(x-1)(x+2)=0$
3. $(2x-1)(x+4)=0$
4. $(x+5)(x-2)=0$

Introduction

Quadratic equations can be a challenging topic for many students and professionals. In this article, we will address some of the most frequently asked questions about quadratic equations, providing clear and concise answers to help you better understand this important mathematical concept.

Q: What is a quadratic equation?

A: A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable (in this case, $x$) is two. The general form of a quadratic equation is:

$$ax^2 + bx + c = 0$$

where $a$, $b$, and $c$ are constants, and $a$ cannot be zero.

Q: How do I solve a quadratic equation?

A: To solve a quadratic equation, you need to follow these steps:

1. Factor the equation: Factor the quadratic expression into two binomial factors.
2. Set each factor equal to zero: Set each binomial factor equal to zero and solve for $x$.
3. Solve for $x$: Solve the resulting linear equations to find the values of $x$.

Q: What is the difference between a quadratic equation and a linear equation?

A: A linear equation is a polynomial equation of degree one, which means the highest power of the variable (in this case, $x$) is one. The general form of a linear equation is:

$$ax + b = 0$$

where $a$ and $b$ are constants, and $a$ cannot be zero.

Q: Can I use the quadratic formula to solve quadratic equations?

A: Yes, you can use the quadratic formula to solve quadratic equations. The quadratic formula is:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a$, $b$, and $c$ are the coefficients of the quadratic equation.

Q: What is the significance of the discriminant in the quadratic formula?

A: The discriminant is the expression under the square root in the quadratic formula:

$$b^2 - 4ac$$

If the discriminant is positive, the quadratic equation has two distinct real solutions. If the discriminant is zero, the quadratic equation has one real solution. If the discriminant is negative, the quadratic equation has no real solutions.

Q: Can I use the quadratic formula to solve quadratic equations with complex solutions?

A: Yes, you can use the quadratic formula to solve quadratic equations with complex solutions. The quadratic formula will give you two complex solutions, which can be written in the form:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a$, $b$, and $c$ are the coefficients of the quadratic equation.

Q: How do I determine the number of solutions to a quadratic equation?

A: To determine the number of solutions to a quadratic equation, you need to examine the discriminant. If the discriminant is:

• Positive, the quadratic equation has two distinct real solutions.
• Zero, the quadratic equation has one real solution.
• Negative, the quadratic equation has no real solutions.

Q: Can I use technology to solve quadratic equations?

A: Yes, you can use technology to solve quadratic equations. Many graphing calculators and computer algebra systems can solve quadratic equations and provide the solutions in the form of a graph or a list of solutions.

Conclusion

In this article, we have addressed some of the most frequently asked questions about quadratic equations. We hope that this article has provided you with a better understanding of this important mathematical concept and has helped you to overcome any challenges you may have faced in solving quadratic equations.